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Finding dimensions of a rectangle. 
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#1
Dec812, 05:19 AM

P: 234

1. The problem statement, all variables and given/known data
the perimeter of a rectangle is 24 ft. the length is 4 ft longer than the width find the dimensions width x length x+4 however, I should be doing it like this: a first equation should start like: 2x+2y=? and the second should start like x=y+? so whats the length and the width? 2. Relevant equations 3. The attempt at a solution x + x + (x+4)+(x+4) = 24 4x+8=24 8 8 4x = 16 4x/4 = 16/4 x = 4 so width 4 ft and length 4+4, so 8 ft. makes sense, since length should be greater than width. 


#3
Dec812, 12:52 PM

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#4
Dec812, 01:21 PM

P: 181

Finding dimensions of a rectangle.
If you've learned one way of solving problems of this kind, and you're expected to show that you have, in fact, understood the technique, then you probably should be doing just what you are, namely translating the facts bit by bit into equations, i.e.
Start with any width you like, say 7. Then the length will be 4 more, so 11. This gives you a perimeter of 7+7+11+11 = 36. Oops, that's 12 too much. Let's subtract this "extra" length from the four sides, taking 12/4 = 3 away from each: width now is 73 = 4, and length is 113 = 8. 


#5
Dec912, 06:51 AM

P: 234

thankyou phinds for confirming.
Thanks for completing those equations Michael Redei, i seen them done that way and I didn't understand it, its much more clear now. 


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